\(\int \frac {-1600 x^2-3200 x^3+e^{2 x} (-400 x^2-800 x^3)+160 \log (4)+e^x (1600 x^2+3200 x^3+(-80-80 x+20 x^2) \log (4))}{1600 x^4+3200 x^5+1600 x^6+e^{2 x} (400 x^4+800 x^5+400 x^6)+(320 x^2+240 x^3-80 x^4) \log (4)+(16-8 x+x^2) \log ^2(4)+e^x (-1600 x^4-3200 x^5-1600 x^6+(-160 x^2-120 x^3+40 x^4) \log (4))} \, dx\) [1976]

Optimal result

Integrand size = 172, antiderivative size = 26 \[ \int \frac {-1600 x^2-3200 x^3+e^{2 x} \left (-400 x^2-800 x^3\right )+160 \log (4)+e^x \left (1600 x^2+3200 x^3+\left (-80-80 x+20 x^2\right ) \log (4)\right )}{1600 x^4+3200 x^5+1600 x^6+e^{2 x} \left (400 x^4+800 x^5+400 x^6\right )+\left (320 x^2+240 x^3-80 x^4\right ) \log (4)+\left (16-8 x+x^2\right ) \log ^2(4)+e^x \left (-1600 x^4-3200 x^5-1600 x^6+\left (-160 x^2-120 x^3+40 x^4\right ) \log (4)\right )} \, dx=\frac {1}{x+x^2+\frac {(-4+x) \log (4)}{20 \left (-2+e^x\right ) x}} \] Output:

1/(x^2+1/10*(-4+x)/x/(exp(x)-2)*ln(2)+x)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {-1600 x^2-3200 x^3+e^{2 x} \left (-400 x^2-800 x^3\right )+160 \log (4)+e^x \left (1600 x^2+3200 x^3+\left (-80-80 x+20 x^2\right ) \log (4)\right )}{1600 x^4+3200 x^5+1600 x^6+e^{2 x} \left (400 x^4+800 x^5+400 x^6\right )+\left (320 x^2+240 x^3-80 x^4\right ) \log (4)+\left (16-8 x+x^2\right ) \log ^2(4)+e^x \left (-1600 x^4-3200 x^5-1600 x^6+\left (-160 x^2-120 x^3+40 x^4\right ) \log (4)\right )} \, dx=\frac {20 \left (-2+e^x\right ) x}{20 \left (-2+e^x\right ) x^2+20 \left (-2+e^x\right ) x^3-4 \log (4)+x \log (4)} \] Input:

Integrate[(-1600*x^2 - 3200*x^3 + E^(2*x)*(-400*x^2 - 800*x^3) + 160*Log[4 
] + E^x*(1600*x^2 + 3200*x^3 + (-80 - 80*x + 20*x^2)*Log[4]))/(1600*x^4 + 
3200*x^5 + 1600*x^6 + E^(2*x)*(400*x^4 + 800*x^5 + 400*x^6) + (320*x^2 + 2 
40*x^3 - 80*x^4)*Log[4] + (16 - 8*x + x^2)*Log[4]^2 + E^x*(-1600*x^4 - 320 
0*x^5 - 1600*x^6 + (-160*x^2 - 120*x^3 + 40*x^4)*Log[4])),x]
 

Output:

(20*(-2 + E^x)*x)/(20*(-2 + E^x)*x^2 + 20*(-2 + E^x)*x^3 - 4*Log[4] + x*Lo 
g[4])
 

Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {-3200 x^3-1600 x^2+e^{2 x} \left (-800 x^3-400 x^2\right )+e^x \left (3200 x^3+1600 x^2+\left (20 x^2-80 x-80\right ) \log (4)\right )+160 \log (4)}{1600 x^6+3200 x^5+1600 x^4+\left (x^2-8 x+16\right ) \log ^2(4)+e^{2 x} \left (400 x^6+800 x^5+400 x^4\right )+\left (-80 x^4+240 x^3+320 x^2\right ) \log (4)+e^x \left (-1600 x^6-3200 x^5-1600 x^4+\left (40 x^4-120 x^3-160 x^2\right ) \log (4)\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {20 \left (-40 \left (e^x-2\right )^2 x^3-x^2 \left (20 e^{2 x}-e^x (80+\log (4))+80\right )-4 e^x x \log (4)-4 \left (e^x-2\right ) \log (4)\right )}{\left (20 \left (e^x-2\right ) x^3+20 \left (e^x-2\right ) x^2+x \log (4)-4 \log (4)\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 20 \int -\frac {40 \left (2-e^x\right )^2 x^3+\left (80+20 e^{2 x}-e^x (80+\log (4))\right ) x^2+4 e^x \log (4) x-4 \left (2-e^x\right ) \log (4)}{\left (20 \left (2-e^x\right ) x^3+20 \left (2-e^x\right ) x^2-\log (4) x+4 \log (4)\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -20 \int \frac {40 \left (2-e^x\right )^2 x^3+\left (80+20 e^{2 x}-e^x (80+\log (4))\right ) x^2+4 e^x \log (4) x-4 \left (2-e^x\right ) \log (4)}{\left (20 \left (2-e^x\right ) x^3+20 \left (2-e^x\right ) x^2-\log (4) x+\log (256)\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -20 \int \left (\frac {2 x+1}{20 x^2 (x+1)^2}+\frac {\log (4) x^3+\log (4) x^2-22 \log (4) x-2 \log (4096)}{20 x^2 (x+1)^2 \left (-20 e^x x^3+40 x^3-20 e^x x^2+40 x^2-\log (4) x+\log (256)\right )}+\frac {-40 \log (4) x^6+80 \log (4) x^5+280 \left (1+\frac {\log (2)}{140}\right ) \log (4) x^4+160 \left (1-\frac {\log (2)}{16}\right ) \log (4) x^3-\log (4) \log (4194304) x^2-4 \log ^2(4) \left (1-\frac {3 \log (256) \log (16384)}{\log (16) \log (64)}\right ) x+2 \log ^2(256)}{20 x^2 (x+1)^2 \left (-20 e^x x^3+40 x^3-20 e^x x^2+40 x^2-\log (4) x+\log (256)\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -20 \left (-\frac {\log (4) \left (312 \log ^3(4)-3 \log ^3(256)+\log (16) \log (64) \log (1024)\right ) \int \frac {1}{(x+1)^2 \left (-20 e^x x^3+40 x^3-20 e^x x^2+40 x^2-\log (4) x+\log (256)\right )^2}dx}{20 \log (16) \log (64)}+\frac {1}{20} \log ^2(4) \int \frac {1}{\left (20 e^x x^3-40 x^3+20 e^x x^2-40 x^2+\log (4) x-\log (256)\right )^2}dx+\frac {1}{10} \log ^2(256) \int \frac {1}{x^2 \left (20 e^x x^3-40 x^3+20 e^x x^2-40 x^2+\log (4) x-\log (256)\right )^2}dx-\frac {\log (2) \left (-8 \log (64) \log ^2(256)-8 \log ^2(4) (\log (64)-6 \log (16384))+7 \log (4) \log (16) \log (64)\right ) \int \frac {1}{(x+1) \left (-20 e^x x^3+40 x^3-20 e^x x^2+40 x^2-\log (4) x+\log (256)\right )^2}dx}{10 \log (16) \log (64)}+\frac {1}{5} \left (\frac {39 \log ^3(4)}{\log (64)}-\log ^2(256)\right ) \int \frac {1}{x \left (-20 e^x x^3+40 x^3-20 e^x x^2+40 x^2-\log (4) x+\log (256)\right )^2}dx+8 \log (4) \int \frac {x}{\left (20 e^x x^3-40 x^3+20 e^x x^2-40 x^2+\log (4) x-\log (256)\right )^2}dx-2 \log (4) \int \frac {x^2}{\left (20 e^x x^3-40 x^3+20 e^x x^2-40 x^2+\log (4) x-\log (256)\right )^2}dx+\frac {1}{10} \log (4096) \int \frac {1}{x^2 \left (20 e^x x^3-40 x^3+20 e^x x^2-40 x^2+\log (4) x-\log (256)\right )}dx-\frac {1}{20} \log (16) \int \frac {1}{x \left (20 e^x x^3-40 x^3+20 e^x x^2-40 x^2+\log (4) x-\log (256)\right )}dx-\frac {1}{20} \log (1048576) \int \frac {1}{(x+1)^2 \left (20 e^x x^3-40 x^3+20 e^x x^2-40 x^2+\log (4) x-\log (256)\right )}dx+\frac {1}{20} \log (4) \int \frac {1}{(x+1) \left (20 e^x x^3-40 x^3+20 e^x x^2-40 x^2+\log (4) x-\log (256)\right )}dx-\frac {1}{20 x (x+1)}\right )\)

Input:

Int[(-1600*x^2 - 3200*x^3 + E^(2*x)*(-400*x^2 - 800*x^3) + 160*Log[4] + E^ 
x*(1600*x^2 + 3200*x^3 + (-80 - 80*x + 20*x^2)*Log[4]))/(1600*x^4 + 3200*x 
^5 + 1600*x^6 + E^(2*x)*(400*x^4 + 800*x^5 + 400*x^6) + (320*x^2 + 240*x^3 
 - 80*x^4)*Log[4] + (16 - 8*x + x^2)*Log[4]^2 + E^x*(-1600*x^4 - 3200*x^5 
- 1600*x^6 + (-160*x^2 - 120*x^3 + 40*x^4)*Log[4])),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77

method	result	size
norman	\(\frac {-20 x +10 \,{\mathrm e}^{x} x}{10 \,{\mathrm e}^{x} x^{3}+10 \,{\mathrm e}^{x} x^{2}-20 x^{3}+x \ln \left (2\right )-20 x^{2}-4 \ln \left (2\right )}\)	\(46\)
parallelrisch	\(\frac {100 \,{\mathrm e}^{x} x -200 x}{100 \,{\mathrm e}^{x} x^{3}+100 \,{\mathrm e}^{x} x^{2}-200 x^{3}+10 x \ln \left (2\right )-200 x^{2}-40 \ln \left (2\right )}\)	\(47\)
risch	\(\frac {1}{\left (1+x \right ) x}-\frac {\left (x -4\right ) \ln \left (2\right )}{\left (1+x \right ) x \left (10 \,{\mathrm e}^{x} x^{3}+10 \,{\mathrm e}^{x} x^{2}-20 x^{3}+x \ln \left (2\right )-20 x^{2}-4 \ln \left (2\right )\right )}\)	\(61\)

Input:

int(((-800*x^3-400*x^2)*exp(x)^2+(2*(20*x^2-80*x-80)*ln(2)+3200*x^3+1600*x 
^2)*exp(x)+320*ln(2)-3200*x^3-1600*x^2)/((400*x^6+800*x^5+400*x^4)*exp(x)^ 
2+(2*(40*x^4-120*x^3-160*x^2)*ln(2)-1600*x^6-3200*x^5-1600*x^4)*exp(x)+4*( 
x^2-8*x+16)*ln(2)^2+2*(-80*x^4+240*x^3+320*x^2)*ln(2)+1600*x^6+3200*x^5+16 
00*x^4),x,method=_RETURNVERBOSE)
 

Output:

(-20*x+10*exp(x)*x)/(10*exp(x)*x^3+10*exp(x)*x^2-20*x^3+x*ln(2)-20*x^2-4*l 
n(2))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {-1600 x^2-3200 x^3+e^{2 x} \left (-400 x^2-800 x^3\right )+160 \log (4)+e^x \left (1600 x^2+3200 x^3+\left (-80-80 x+20 x^2\right ) \log (4)\right )}{1600 x^4+3200 x^5+1600 x^6+e^{2 x} \left (400 x^4+800 x^5+400 x^6\right )+\left (320 x^2+240 x^3-80 x^4\right ) \log (4)+\left (16-8 x+x^2\right ) \log ^2(4)+e^x \left (-1600 x^4-3200 x^5-1600 x^6+\left (-160 x^2-120 x^3+40 x^4\right ) \log (4)\right )} \, dx=-\frac {10 \, {\left (x e^{x} - 2 \, x\right )}}{20 \, x^{3} + 20 \, x^{2} - 10 \, {\left (x^{3} + x^{2}\right )} e^{x} - {\left (x - 4\right )} \log \left (2\right )} \] Input:

integrate(((-800*x^3-400*x^2)*exp(x)^2+(2*(20*x^2-80*x-80)*log(2)+3200*x^3 
+1600*x^2)*exp(x)+320*log(2)-3200*x^3-1600*x^2)/((400*x^6+800*x^5+400*x^4) 
*exp(x)^2+(2*(40*x^4-120*x^3-160*x^2)*log(2)-1600*x^6-3200*x^5-1600*x^4)*e 
xp(x)+4*(x^2-8*x+16)*log(2)^2+2*(-80*x^4+240*x^3+320*x^2)*log(2)+1600*x^6+ 
3200*x^5+1600*x^4),x, algorithm="fricas")
 

Output:

-10*(x*e^x - 2*x)/(20*x^3 + 20*x^2 - 10*(x^3 + x^2)*e^x - (x - 4)*log(2))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (22) = 44\).

Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.73 \[ \int \frac {-1600 x^2-3200 x^3+e^{2 x} \left (-400 x^2-800 x^3\right )+160 \log (4)+e^x \left (1600 x^2+3200 x^3+\left (-80-80 x+20 x^2\right ) \log (4)\right )}{1600 x^4+3200 x^5+1600 x^6+e^{2 x} \left (400 x^4+800 x^5+400 x^6\right )+\left (320 x^2+240 x^3-80 x^4\right ) \log (4)+\left (16-8 x+x^2\right ) \log ^2(4)+e^x \left (-1600 x^4-3200 x^5-1600 x^6+\left (-160 x^2-120 x^3+40 x^4\right ) \log (4)\right )} \, dx=\frac {- x \log {\left (2 \right )} + 4 \log {\left (2 \right )}}{- 20 x^{5} - 40 x^{4} - 20 x^{3} + x^{3} \log {\left (2 \right )} - 3 x^{2} \log {\left (2 \right )} - 4 x \log {\left (2 \right )} + \left (10 x^{5} + 20 x^{4} + 10 x^{3}\right ) e^{x}} + \frac {1}{x^{2} + x} \] Input:

integrate(((-800*x**3-400*x**2)*exp(x)**2+(2*(20*x**2-80*x-80)*ln(2)+3200* 
x**3+1600*x**2)*exp(x)+320*ln(2)-3200*x**3-1600*x**2)/((400*x**6+800*x**5+ 
400*x**4)*exp(x)**2+(2*(40*x**4-120*x**3-160*x**2)*ln(2)-1600*x**6-3200*x* 
*5-1600*x**4)*exp(x)+4*(x**2-8*x+16)*ln(2)**2+2*(-80*x**4+240*x**3+320*x** 
2)*ln(2)+1600*x**6+3200*x**5+1600*x**4),x)
 

Output:

(-x*log(2) + 4*log(2))/(-20*x**5 - 40*x**4 - 20*x**3 + x**3*log(2) - 3*x** 
2*log(2) - 4*x*log(2) + (10*x**5 + 20*x**4 + 10*x**3)*exp(x)) + 1/(x**2 + 
x)
 

Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {-1600 x^2-3200 x^3+e^{2 x} \left (-400 x^2-800 x^3\right )+160 \log (4)+e^x \left (1600 x^2+3200 x^3+\left (-80-80 x+20 x^2\right ) \log (4)\right )}{1600 x^4+3200 x^5+1600 x^6+e^{2 x} \left (400 x^4+800 x^5+400 x^6\right )+\left (320 x^2+240 x^3-80 x^4\right ) \log (4)+\left (16-8 x+x^2\right ) \log ^2(4)+e^x \left (-1600 x^4-3200 x^5-1600 x^6+\left (-160 x^2-120 x^3+40 x^4\right ) \log (4)\right )} \, dx=-\frac {10 \, {\left (x e^{x} - 2 \, x\right )}}{20 \, x^{3} + 20 \, x^{2} - 10 \, {\left (x^{3} + x^{2}\right )} e^{x} - x \log \left (2\right ) + 4 \, \log \left (2\right )} \] Input:

integrate(((-800*x^3-400*x^2)*exp(x)^2+(2*(20*x^2-80*x-80)*log(2)+3200*x^3 
+1600*x^2)*exp(x)+320*log(2)-3200*x^3-1600*x^2)/((400*x^6+800*x^5+400*x^4) 
*exp(x)^2+(2*(40*x^4-120*x^3-160*x^2)*log(2)-1600*x^6-3200*x^5-1600*x^4)*e 
xp(x)+4*(x^2-8*x+16)*log(2)^2+2*(-80*x^4+240*x^3+320*x^2)*log(2)+1600*x^6+ 
3200*x^5+1600*x^4),x, algorithm="maxima")
 

Output:

-10*(x*e^x - 2*x)/(20*x^3 + 20*x^2 - 10*(x^3 + x^2)*e^x - x*log(2) + 4*log 
(2))
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {-1600 x^2-3200 x^3+e^{2 x} \left (-400 x^2-800 x^3\right )+160 \log (4)+e^x \left (1600 x^2+3200 x^3+\left (-80-80 x+20 x^2\right ) \log (4)\right )}{1600 x^4+3200 x^5+1600 x^6+e^{2 x} \left (400 x^4+800 x^5+400 x^6\right )+\left (320 x^2+240 x^3-80 x^4\right ) \log (4)+\left (16-8 x+x^2\right ) \log ^2(4)+e^x \left (-1600 x^4-3200 x^5-1600 x^6+\left (-160 x^2-120 x^3+40 x^4\right ) \log (4)\right )} \, dx=\frac {10 \, {\left (x e^{x} - 2 \, x\right )}}{10 \, x^{3} e^{x} - 20 \, x^{3} + 10 \, x^{2} e^{x} - 20 \, x^{2} + x \log \left (2\right ) - 4 \, \log \left (2\right )} \] Input:

integrate(((-800*x^3-400*x^2)*exp(x)^2+(2*(20*x^2-80*x-80)*log(2)+3200*x^3 
+1600*x^2)*exp(x)+320*log(2)-3200*x^3-1600*x^2)/((400*x^6+800*x^5+400*x^4) 
*exp(x)^2+(2*(40*x^4-120*x^3-160*x^2)*log(2)-1600*x^6-3200*x^5-1600*x^4)*e 
xp(x)+4*(x^2-8*x+16)*log(2)^2+2*(-80*x^4+240*x^3+320*x^2)*log(2)+1600*x^6+ 
3200*x^5+1600*x^4),x, algorithm="giac")
 

Output:

10*(x*e^x - 2*x)/(10*x^3*e^x - 20*x^3 + 10*x^2*e^x - 20*x^2 + x*log(2) - 4 
*log(2))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-1600 x^2-3200 x^3+e^{2 x} \left (-400 x^2-800 x^3\right )+160 \log (4)+e^x \left (1600 x^2+3200 x^3+\left (-80-80 x+20 x^2\right ) \log (4)\right )}{1600 x^4+3200 x^5+1600 x^6+e^{2 x} \left (400 x^4+800 x^5+400 x^6\right )+\left (320 x^2+240 x^3-80 x^4\right ) \log (4)+\left (16-8 x+x^2\right ) \log ^2(4)+e^x \left (-1600 x^4-3200 x^5-1600 x^6+\left (-160 x^2-120 x^3+40 x^4\right ) \log (4)\right )} \, dx=\int -\frac {{\mathrm {e}}^{2\,x}\,\left (800\,x^3+400\,x^2\right )-{\mathrm {e}}^x\,\left (1600\,x^2-2\,\ln \left (2\right )\,\left (-20\,x^2+80\,x+80\right )+3200\,x^3\right )-320\,\ln \left (2\right )+1600\,x^2+3200\,x^3}{4\,{\ln \left (2\right )}^2\,\left (x^2-8\,x+16\right )-{\mathrm {e}}^x\,\left (2\,\ln \left (2\right )\,\left (-40\,x^4+120\,x^3+160\,x^2\right )+1600\,x^4+3200\,x^5+1600\,x^6\right )+{\mathrm {e}}^{2\,x}\,\left (400\,x^6+800\,x^5+400\,x^4\right )+2\,\ln \left (2\right )\,\left (-80\,x^4+240\,x^3+320\,x^2\right )+1600\,x^4+3200\,x^5+1600\,x^6} \,d x \] Input:

int(-(exp(2*x)*(400*x^2 + 800*x^3) - exp(x)*(1600*x^2 - 2*log(2)*(80*x - 2 
0*x^2 + 80) + 3200*x^3) - 320*log(2) + 1600*x^2 + 3200*x^3)/(4*log(2)^2*(x 
^2 - 8*x + 16) - exp(x)*(2*log(2)*(160*x^2 + 120*x^3 - 40*x^4) + 1600*x^4 
+ 3200*x^5 + 1600*x^6) + exp(2*x)*(400*x^4 + 800*x^5 + 400*x^6) + 2*log(2) 
*(320*x^2 + 240*x^3 - 80*x^4) + 1600*x^4 + 3200*x^5 + 1600*x^6),x)
 

Output:

int(-(exp(2*x)*(400*x^2 + 800*x^3) - exp(x)*(1600*x^2 - 2*log(2)*(80*x - 2 
0*x^2 + 80) + 3200*x^3) - 320*log(2) + 1600*x^2 + 3200*x^3)/(4*log(2)^2*(x 
^2 - 8*x + 16) - exp(x)*(2*log(2)*(160*x^2 + 120*x^3 - 40*x^4) + 1600*x^4 
+ 3200*x^5 + 1600*x^6) + exp(2*x)*(400*x^4 + 800*x^5 + 400*x^6) + 2*log(2) 
*(320*x^2 + 240*x^3 - 80*x^4) + 1600*x^4 + 3200*x^5 + 1600*x^6), x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {-1600 x^2-3200 x^3+e^{2 x} \left (-400 x^2-800 x^3\right )+160 \log (4)+e^x \left (1600 x^2+3200 x^3+\left (-80-80 x+20 x^2\right ) \log (4)\right )}{1600 x^4+3200 x^5+1600 x^6+e^{2 x} \left (400 x^4+800 x^5+400 x^6\right )+\left (320 x^2+240 x^3-80 x^4\right ) \log (4)+\left (16-8 x+x^2\right ) \log ^2(4)+e^x \left (-1600 x^4-3200 x^5-1600 x^6+\left (-160 x^2-120 x^3+40 x^4\right ) \log (4)\right )} \, dx=\frac {10 x \left (e^{x}-2\right )}{10 e^{x} x^{3}+10 e^{x} x^{2}+\mathrm {log}\left (2\right ) x -4 \,\mathrm {log}\left (2\right )-20 x^{3}-20 x^{2}} \] Input:

int(((-800*x^3-400*x^2)*exp(x)^2+(2*(20*x^2-80*x-80)*log(2)+3200*x^3+1600* 
x^2)*exp(x)+320*log(2)-3200*x^3-1600*x^2)/((400*x^6+800*x^5+400*x^4)*exp(x 
)^2+(2*(40*x^4-120*x^3-160*x^2)*log(2)-1600*x^6-3200*x^5-1600*x^4)*exp(x)+ 
4*(x^2-8*x+16)*log(2)^2+2*(-80*x^4+240*x^3+320*x^2)*log(2)+1600*x^6+3200*x 
^5+1600*x^4),x)
 

Output:

(10*x*(e**x - 2))/(10*e**x*x**3 + 10*e**x*x**2 + log(2)*x - 4*log(2) - 20* 
x**3 - 20*x**2)